How to show right continuity for a df? Suppose $\mu$ is a Lebesgue Stieltjes measure on $(\mathbb{R},B_{\mathbb{R}})$. Define the function $F$ as follows: 
$F(x)=\begin{cases} 
      \mu((0,x]) & x > 0 \\
      0 & x=0\\
      - \mu ((x,0]) & x<0 \\
   \end{cases}
$
We need to show that $F$ is a distribution function. 
I have shown that $F$ is non-decreasing. 
All I need to show is that $F$ is right continuous. That is now if we take a sequence $s_n \rightarrow 0^{+}$, we need to show that suppose for $x>0$, $\mu((0,x+s_n]) \rightarrow \mu((0,x])$. This is trivial if $s_n$ is a monotone sequence as we can directly use continuity from above. But $s_n$ here can be any arbitrary sequence going to $0$. So, how can we prove this?
 A: I will assume that $\mu (-N,N) <\infty$ for all $N$. 
This condition is satisfied by any L-S measure: see https://www.oreilly.com/library/view/measure-probability-and/9781118831984/OEBPS/c04.htm
for the definition.
For $x>0$, $\mu((0,x+s_n]) = \mu((0,x]) +\mu((x,x+s_n])$  and the intervals $(x,x+s_n]$ decrease to the empty set as $s_n$ decreases to $0$. The proof for $x<0$ is similar.
Let $s_n \to 0+$ and $\epsilon >0$. Let $m=\{\inf f(y): y >x\}$. Then $ m \leq f(x+s_n )$ for all $n$. Next let $\epsilon>0$ and note that there exists $y>x$ such that $f(y) <m+\epsilon$. Now $f(x+s_n) \leq  f(y) <m+\epsilon$ for  $n$ sufficiently large by monotonicity of $f$. If you combine the two facts we just proves you can see that $f(x+s_n) \to m$ for any sequence $s_n \to 0+$. 
A: It suffices to consider for $s_n\searrow0$ and show that $\lim_nF(x+s_n)=F(x)$.
$$\bigcap_n(0,x+s_n]= (0,x]$$
for all $x\geq0$. 
$$ \bigcup_n(x+s_n,0]=(x,0]$$
for all $x<0$.
Use the fact that $\mu$ is a Lebeusgue-Stieltjes measure, which imply that $\mu$ is finite in bounded sets. The rest is a well known result that says that $\mu(A_n)\nearrow\mu(A)$ when $A_n\nearrow A$ and $\mu(A_n)\searrow\mu(A)$ when $\mu(A_1)<\infty$ and $A_n\searrow A$.
